Dane Sabo
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100 lines
3.6 KiB
TeX
100 lines
3.6 KiB
TeX
\section*{Problem 7}
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\subsection*{Part A}
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Starting with the point kinetics equations with one delayed precursor group, using
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the prompt jump approximation where prompt neutrons reach equilibrium instantly:
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\[\frac{dN(t)}{dt} = \lambda C(t)\]
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\[\frac{dC(t)}{dt} = \frac{\beta}{\Lambda}N(t) - \lambda C(t)\]
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Using appropriate numerical values for a PWR with low enrichment U-235 fuel:
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\begin{itemize}
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\item \(\beta = 0.0065\) (650 pcm)
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\item \(\Lambda = 5 \times 10^{-5}\) s (50 \(\mu\)s)
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\item \(\lambda = 0.08\) s\(^{-1}\)
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\item \(\rho = +0.0005\) (+50 pcm inserted)
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\end{itemize}
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The complete system of equations describing reactor power as a function of time is:
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\[N(t) = \Lambda \frac{\lambda C(t)}{\beta -\rho} \]
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\[\frac{dC(t)}{dt} = \frac{\beta}{\Lambda}N(t) - \lambda C(t)\]
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Substituting N(t) into the precursor equation:
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\[\frac{dC(t)}{dt} = \frac{\beta \lambda}{\beta -\rho} C(t) - \lambda C(t) = \lambda C(t) \left(\frac{\beta}{\beta - \rho} - 1\right)\]
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Simplifying:
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\[\frac{dC(t)}{dt} = \lambda C(t) \frac{\rho}{\beta - \rho}\]
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With the given numerical values (\(\beta = 0.0065\), \(\rho = 0.0005\), \(\lambda = 0.08\) s\(^{-1}\), \(\Lambda = 5 \times 10^{-5}\) s):
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\[\frac{dC(t)}{dt} = 0.08 \times C(t) \times \frac{0.0005}{0.006} = 6.67 \times 10^{-3} C(t)\]
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\[N(t) = 5 \times 10^{-5} \times \frac{0.08 \times C(t)}{0.006} = 6.67 \times 10^{-4} C(t)\]
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\subsection*{Part B}
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The kinetics parameters chosen are justified as follows:
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\begin{itemize}
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\item \(\beta = 0.0065\): This is the standard delayed neutron fraction for
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thermal fission of U-235.
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\item \(\Lambda = 50 \mu\)s: This is the typical prompt neutron generation
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time for a pressurized water reactor.
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\item \(\lambda = 0.08\) s\(^{-1}\): This effective decay constant represents
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the lumping of six delayed neutron precursor groups into one equivalent group.
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This gives an effective half-life of \(t_{1/2} = \ln(2)/\lambda \approx 8.7\)
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s, which is reasonable as a weighted average of the six precursor groups that
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range from \(\sim\)0.2 s to \(\sim\)80 s half-lives.
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\end{itemize}
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\subsection*{Part C}
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Power will stop increasing when \(\frac{dN}{dt} = 0\), which occurs when
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temperature feedback effects add sufficient negative reactivity to cancel the
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+50 pcm reactivity insertion.
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The reactor reaches a new equilibrium when:
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\[\rho_{net} = \rho_{inserted} + \alpha_f \Delta T_f + \alpha_m \Delta T_m = 0\]
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For \(\rho_{inserted} = +50\) pcm, temperature feedback must provide
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\(-50\) pcm.
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\subsection*{Part D}
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At end of life (EOL) with +200 pcm reactivity insertion, several key parameters
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change:
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\begin{itemize}
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\item \(\beta_{EOL} \approx 0.005\) (500 pcm): At EOL, approximately 40-50\%
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of fissions come from Pu-239 and Pu-241 built up from neutron capture in
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U-238. Since Pu-239 has \(\beta \approx 0.0021\) (much lower than U-235's
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0.0065), the effective \(\beta\) is a weighted average that decreases to
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around 0.005.
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\end{itemize}
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The time constant for the transient becomes:
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\[\tau_{EOL} = \frac{\beta - \rho}{\lambda \rho} = \frac{0.0045 - 0.002}{0.08 \times 0.002} = 15.6 \text{ s}\]
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compared to BOL:
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\[\tau_{BOL} = \frac{0.0065 - 0.0005}{0.08 \times 0.0005} = 150 \text{ s}\]
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The EOL transient is much faster (10x shorter period) because \(\beta - \rho\)
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is smaller relative to \(\rho\).
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The spent core actually experiences a more dramatic transient per unit
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reactivity insertion, but the stronger feedback from increased fuel temperature
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effects provides self-limiting behavior that prevents excessive power excursion.
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Plutonium-240 absorbs a lot of neutrons as 240Pu concentration builds up.
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